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In past weeks, the average number of life insurance policies sold per week was $θ = 3$. However, this week he has sold 6 life insurance policies. Based on this single observation of $x = 6$, what is the value of the likelihood-ratio test statistic for testing the hypothesis that the salesman has suddenly seen a boost in his job, i.e. testing $H_0 = \theta = 3$ , $H_1 = \theta > 3$.

For this, I used the likelihood ratio for poisson distribution, giving me $$\frac{\frac{3^6e^{-3}}{6!}}{\frac{6^6e^{-6}}{6!}}$$ This gives 0.3138365. Is this the right answer?

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  • $\begingroup$ Your fraction reduces to $3^3/6^6 = .5^6= 0.015625.$ $\endgroup$
    – BruceET
    Apr 29, 2019 at 19:52
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    $\begingroup$ @BruceET - thought the fraction was probably intended to be $\frac{\frac{3^6 e^{-3}}{6!}}{\frac{6^6e^{-6}}{6!}} =\dfrac{\exp(3)}{2^6}$ $\endgroup$
    – Henry
    Apr 29, 2019 at 20:01
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    $\begingroup$ Yes that was the intended fraction, just typed it wrong. Thank you for clarifying! $\endgroup$ Apr 29, 2019 at 20:15

1 Answer 1

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You will want to reject $H_0: \theta = 3$ against $H_a: \theta > 3)$ for large values of your single observation $X.$ You can't test at exactly the 5% level because of the discreteness of the Poisson distribution.

Under $H_0,$ you have $P(X \ge 7) = .0335,$ but $P(X \ge 6) = .0834.$ Exact Poisson computations from R (where ppois is a Poisson CDF and qpois is the quantile function):

qpois(.95, 3)
[1] 6
1 - ppois(6,3)
[1] 0.03350854  # 1 - P(X <= 6) = P(X >= 7)
1 - ppois(5,3)
[1] 0.08391794

So for test at level $\alpha =3.335,$ you should reject if $X \ge 7.$ Because you observe $X = 6,$ the P-value is $P(X \ge 6\,|\,\theta = 3) = 0.834.$

In the plot below, $\alpha$ is the sum of the heights of the bars to the right of the vertical dotted line.

enter image description here

For such a small values of $\theta,$ depending on the desired accuracy, it may no bet advisable to use a normal approximation. The best-fitting normal curve is shown in blue.

However, using the normal approximation (with continuity correction), the P-value is 7.4%, so you would not reject at the 5% level.

1 - pnorm(5.5, 3, sqrt(3))
[1] 0.07445734

Addendum: Now that you have the correct likelihood ratio function $\lambda(x) = (3/x)^xe^{x-3}$ for $x \ge 3 \; (1$ for $x < 3),$ let's use R to plot it. Then we can see that $\lambda(x)$ is small (leading to rejection) for large values of $x$ as mentioned at the start.

x = 3:10;  like = (3/x)^x*exp(x - 3)
plot(x, like, pch=20, main="Likelihood Ratio")
abline(h = 0, col="green2")

enter image description here

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